# Tag Info

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Let $v = u^4$ and rewrite the PDE in terms of $v$ to get: $$v^{-3/4} v_t = \Delta v.$$ Let $\;v = A(t) B(x,y,z)\;$ and separate variables to get: $$A' = 4 A^{7/4} c \quad \mbox{and} \quad B^{3/4} \Delta B = c \quad \mbox{for a constant } c .$$ The DE in $A$ has the solution $\;A(t)^{3/4} = -\frac{4/3}{4 c t + k}\;$ for a constant $k$. So when $t$ goes to ...

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Numerical Partial Differential Equations I and II by Thomas have become my go-to references. They cover all the major topics (FD, FE and FV) and have lots of exercises.

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The closure of a bounded subset of $\mathbb R^n$ is compact. A continuous function on a compact set is bounded.

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Don't have time to write a long answer, but the short answer is because of boundary conditions. That is, in your first list of equations, you have the boundary condition $u(t,0)=0$. You don't have this boundary condition for the equations involving $w$. However, because you've chosen the initial conditions of $w$ carefully, the initial conditions you are ...

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The sum $\Sigma(\cdots)$ is an asymptotic series as $t \to \infty$, so the first two terms of the expansion of $u(x,t)$ as $t \to \infty$ are simply the two terms which do not decrease exponentially, namely $$u(x,t) \sim \underbrace{4t}_\text{first} + \underbrace{2x-10}_\text{second} + \cdots$$ as $t \to \infty$. Now, the first two terms of the ...

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Let's use the area form of the mean value property: For any $r>0,$ $$|u(0)| = |\frac{1}{\pi r^2}\int_{D(0,r)}u\,dx\,dy\,| \le \frac{1}{\pi r^2}\int_{D(0,r)}|u|\,dx\,dy$$ $$\le \frac{1}{\pi r^2}\int_r^r\int_r^r|u(x,y)|\,dx\,dy\le \frac{1}{\pi r^2}\cdot C\cdot 2r = \frac{2C}{\pi r}.$$ Let $r\to \infty$ to see $u(0)=0.$ Clearly we could repeat the ...

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I found the solution: Since $\bigtriangleup u = 2x(y −1)(y −2x + x y +2)e^{x−y}$, $(x, y) ∈ (0, 1)×(0, 1)$ The function u(x,t) must have $e^{x−y}$ From the boundary conditions: $u(x, 0) = 0$ then u(x,t) must be 0 for x=0, so we assume u(x,t) must have $e^{x−y}*x$ $u(x, 1) = 0$ then u(x,t) must be 0 for x=1, so we assume u(x,t) must have ...

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This book is very good: Numerical Solution of Partial Differential Equations Morton and Mayers

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This is really an ODE, it suffices to make two arbitrary integration constants in the general solution arbitrary functions of $x$. To solve this ODE, rewrite it as $$\frac{f_{yy}}{f_y}=\frac12\frac{f_y}{f}\qquad \left(\ln f_y\right)_y=\frac12 \left(\ln f\right)_y,$$ which gives $\ln f_y=\frac12\ln f+ \mathrm{const}$ or, in other words, $f_y=2C_1\sqrt f$. ...

2

Suppose that that $\{v_j\}$ are eigenvectors with distinct eigenvalues $\{\lambda_j\}$ and that $$\sum_{j=1}^na_jv_j=0\tag{1}$$ For any $k$ we have \begin{align} 0 &=T^k\sum_{j=1}^na_jv_j\\ &=\sum_{j=1}^na_j\lambda_j^kv_j\tag{2} \end{align} For $0\le k\le n-1$, $(2)$ can be rewritten as $$0=\begin{bmatrix} 1&1&1&\cdots&1\\ ... 1 Let v(x) = x_1^2 + \cdots +x_n^2. Then v \ge 0 and \Delta v = 2n. If u \ge 0 and \Delta u = -1 then u + v/2n \ge 0 and \Delta(u + v/2n) = 0. A nonnegative harmonic function is constant, so that u + v/2n = C for some constant C. This leads rather quickly to a contradiction since v(x) \to \infty as |x| \to \infty. 1 The general form of a harmonic function in an annulus is$$ u(r,\theta) = (A_0 +B_0\log r)+ \sum_{n\ne 0} r^n (A_n \cos |n|\theta+B_n\sin |n|\theta)\tag{1} $$The stated boundary conditions are:$$ 0 = (A_0 +B_0\log 2)+ \sum_{n\ne 0} 2^n (A_n \cos |n|\theta+B_n\sin |n|\theta) \tag{2}$$and$$ \sin\theta = (A_0 +B_0\log 4)+ \sum_{n\ne 0} 4^n (A_n \cos ...

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You have to use the chain rule in the last step. Putting in some extra parentheses for emphasis, you have: $$D_t (\phi(x-ct)) = (\nabla \phi)(x-ct) \cdot (-c)$$ which is a scalar.

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$$f_yf \left(\frac{f_y}{f} -2\frac{f_{yy}}{f_y}\right) = 0$$ Thus we can obtain $$\dfrac{\partial}{\partial y}\ln f -2\dfrac{\partial }{\partial y}\ln f_y = 0$$ Or $$\ln \left(\frac{f}{f_y^2}\right) = g(x)$$ So you get $$f = g_2(x) f_y^2\implies f_y = \sqrt{\frac{f}{g_2(x)}}$$

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Let $y(x,t) = w(x,t) + \phi(x)$ where $\phi(x) = a x^2 + b x + c$. From this it is seen that $\phi'(x) = 2 a x + b$, $\phi''(x) = 2a$ and \begin{align} \partial_{t}^{2} w = c^2 \, \partial_{x}^{2} w + 2a c^2 + L. \end{align} In order to cancel the $L$ term let $2 a c^2 + L = 0$ which leads to \begin{align} \phi(x) = - \frac{L \, x^{2}}{2 \, c^{2}} + b \, x + ...

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Your question is incomplete, one need to read page 46 of Sneddon's book to figure out what the problem really want. To summarize, what the book want is start from an equation of the form $$z = f(u)\quad\text{ where }\quad u = \frac{xy}{z}$$ derive a PDE for $z$ which doesn't involve the function $f(u)$ explicitly. The tool you need is chain rule for ...

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Note that the solution of $-\Delta u = f$, with homogeneous dirichlet BC's is given by $$u(x)=\Phi*f=\int_\Omega\Phi(x-y)f(y)\,dy$$ where $\Phi$ is the fundamental solution of the Laplace equation, and (in 2D) $$\Phi(x)=\frac{\ln(|x|)}{2\pi}.$$ Note that in the sense of distributions, $-\Delta\Phi(x)=\delta(x)$, and so $$-\Delta u = ... 1 you are using the variable t in two different way. you are also forgetting that a solution of \frac{dx}{d\tau} = -y is not x = -\tau y because y is also a function of \tau. i will use a new variable s instead. so we have$$\frac{dt}{ds} = t^2, \frac{dx}{ds} = -y, \frac{dy}{ds} = x \to \frac{d^2x}{ds^2} + x = 0 $$so that$$x = a\cos(s-b), y = ...

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Differentiating the first equation with respect to $x$:$$\frac{\partial u}{\partial t} +g\frac{\partial \eta}{\partial x}=0\implies\frac{\partial^2 u}{\partial t\partial x}+g\frac{\partial^2 \eta}{\partial x^2}=0$$ and the second with respect to $t$: $$\frac{\partial\eta}{\partial t}+H\frac{\partial u}{\partial x}=0\implies \frac{\partial^2\eta}{\partial ... 1$$\frac{\partial u}{\partial t}+g\frac{\partial \eta}{\partial x}=0\frac{\partial\eta}{\partial t}+H\frac{\partial u}{\partial x}=0$$This implies that$$ \frac{\partial}{\partial x}\left(\frac{\partial u}{\partial t}+g\frac{\partial \eta}{\partial x}\right)= \frac{\partial^2 u}{\partial t\partial x}+g\frac{\partial^2 \eta}{\partial x^2}=0$$And$$ ...

1

the characteristics are given by $$\frac{dx}{dt} = x, \frac{dy}{dt} = y, x(0) = 1, y(0) = b$$ which gives $$x = e^t, y = be^t, b = \frac yx, t=\ln x.$$ along the characteristics we have $$\frac{du}{dt} = u(1+b)e^t, u = 1, t=0 \to \int_1^u\frac{du}{u} = (1+b)\int_0^t e^t\, dt$$ integrating we get $$\ln u=(1+b)(e^t-1). u(x, y) = ... 1 You are almost there: By choosing v = 0, we see that L^*(p) \geq -L(0) for all p \in \mathbb{R}^n. The superlinearity of L implies that for each p\in \mathbb{R}^n there is R > 0 such that for all v \in \mathbb{R}^n with |v| > R, $$p \cdot v - L(v) < -L(0).$$ We can therefore safely ignore any v with ... 1 In this case, it helps to solve the problem by solving for \tilde{\phi} = \phi - V. Write the solution as an expansion of the eigenfunctions of the spatial Laplace operator. i.e. Once you use separation of variables, you should be able to see that$$ \tilde{\phi}(r,\theta) = A_o \ln r + B_o + \sum\limits_{n=1}^{\infty} (A_n \cos(n \theta) + B_n \sin(n ...

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You should be looking for $f(x)=g(\|x\|)$, there shouldn't be a squared term, this is because the Laplace operator is rotation invariant, so you can seek a radial solution. Now applying $\Delta$ to $g$, you will end up with an ODE in the variable $r=\|x\|$, which you can solve to find the fundamental solution of the Laplace equation. Note that ...

1

I am starting from $$\sum_{k=1}^\infty c_k \sin \left ( \frac{\pi k x}{l} \right ) \left ( 1 - e^{\frac{2 k \pi^2}{l}} \right ) = f(x).$$ I didn't actually check your work to this point, so be sure to check that. From there, the left side is in the form of a Fourier sine series for $f$. (Note that $f$ depends on $x$, which lives in $[0,l]$, not in ...

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The method of characteristics is appropriate to solve initial value problems of hyperbolic type: semi linear first order differential equations, one-dimensional wave equation. In principle all solutions can be found using this method. Similarity solutions are a special type of solutions that reflect invariant properties of the equation. Very often they ...

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